Question

Find the equation of locus of a point p such that distance of p from the origin is twice the distance of p from a (1,2) solution .

Answer 1

$\text{Let P(h,k) be the moving point and A be (1,2)}$

$\textbf{Given:}$

$OP=2{\times}PA$

$\textbf{To find:}$

$\text{Locus of P}$

$\textbf{Solution:}$

$\text{Consider,}$

$OP=2{\times}PA$

$\sqrt{(0-h)^2+(0-k)^2}=2\sqrt{(h-1)^2+(k-2)^2}$

$\sqrt{h^2+k^2}=2\sqrt{(h-1)^2+(k-2)^2}$

$\text{Squaring on bothsides, we get}$

$h^2+k^2=4[(h-1)^2+(k-2)^2]$

$h^2+k^2=4[h^2+1-2h+k^2+4-4k]$

$h^2+k^2=4[h^2-2h+k^2+5-4k]$

$h^2+k^2=4h^2-8h+4k^2+20-16k$

$\text{Rearranging terms, we get}$

$3h^2+3k^2-8h-16k+20=0$

$\therefore\text{The locus of P is}$

$\boxed{\bf\,3x^2+3y^2-8x-16y+20=0}$

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